help-0

[torpid mirage]

The negative can't come from the sqrt

[uncut tide]

yeah

[torpid mirage]

No we are in a < 0

In the other cases it's positive over the whole domain

[uncut tide]

oh yeah sorry meant to put x

x>0 x<1/a

[torpid mirage]

So when x < 1/b

Or -1/a

Don't forget the -ve

The subtracted part will always be less than one

[uncut tide]

, the asymtote is at 1/a

[torpid mirage]

You can check it out

It will lead to the function always being positive in that part

[uncut tide]

ok

makes sense

[torpid mirage]

-1/a

Or 1/b

a is negative. Remember

[uncut tide]

ahhh

yeah

cool

yeah i see how its always positive

so we have a curve sort of approaching infinty as it goes towards 0

and when it goes towards 1/b

[torpid mirage]

And you see how it's always negative for $x > \frac{1}{b}$ ?

[ocean seal]

Mohamed Mohsen

[uncut tide]

yeah i do

makes sense

[torpid mirage]

I think you are ready to graph it then

[uncut tide]

case 2 one

approches zero from left

we have it never being negative in the middle

approching positive infinity from lhs as x approches 1/b

and negative infinty from rhs

[torpid mirage]

Yes

And the plot is correct but what is that line when x is below zero

You drew a curve when x was less than zero

[uncut tide]

apprching zero from left

[torpid mirage]

Sqrt of negative ?

[uncut tide]

so do we just ingore that then?

[torpid mirage]

The function has no real value for negative x

[uncut tide]

ok so we ignore what we did in a then

makes sense

[torpid mirage]

What did we do in a ?

[uncut tide]

large argument behaviour

sorry b

[torpid mirage]

It was for positive large behavior

Not negative one

We always assumed x was positive

[uncut tide]

oh, yeah makes sense

[torpid mirage]

Negative is not with us in the entire question

[uncut tide]

but then, how does it apprach from the left if positive

[torpid mirage]

It doesn't

It's a one sided lim and a one sided asymptote

[uncut tide]

ok so, if i get a positve large x approching from left, its wrong, like it cant exist

[torpid mirage]

I don't understand

Limits to +inf are always approaching from the left

There is no right side to +inf

[uncut tide]

huh

apprching left from zero

[torpid mirage]

Yes approaching zero is from the right here

[uncut tide]

this is what i have written down

for b

for case 2

it is the only one with an extra asymptote and approaching zero from lhs

like we said, but if x cant be negative this dsnt make sense

[torpid mirage]

In large arg behav notice you said infinity

And not -inf

[uncut tide]

ok

[torpid mirage]

All of this is fine

And nothing implies values for f in the negative region

[uncut tide]

im still a little confused then, appraching zero, from left , means negative values

[torpid mirage]

Yea so you can't do it

In small argument approximation you can specify that x > 0

[uncut tide]

ok so basiaclly i need to write, cant do this as we cant have negtaive values

[torpid mirage]

So this is calculated in the sense of a rhs lim

Yes

[uncut tide]

ok, so the limit we calcuated is essentially invalid as we cant have -x

because the limit implies negative values

[torpid mirage]

You mean in c ?

[uncut tide]

no for be

b

when it says appraching zero from lhs

for case 2

this doesnt work as it means -x values

[torpid mirage]

What no

Look

We have to pay attention to notation here

Because you arw getting confused

What is approaching zero

[uncut tide]

the fucntion

[torpid mirage]

X can't approach zero yes. But is that x we are talking about

[uncut tide]

at large x

[torpid mirage]

Yea

So it can be negative

And we have shown it is negative in case 2

[uncut tide]

ohhhhhhhhh

[torpid mirage]

For x > 1/b

But x can't be negative

[uncut tide]

makes sense

what ur saying is, the function can be negative for large x

[torpid mirage]

Yes

[uncut tide]

ahhhh i get u

[torpid mirage]

But you can't calculate limit as 'x' goes to zero from lhs

[uncut tide]

ahh

[torpid mirage]

In part a we did small argument which is behaviour of f near zero

I recommend you specify that x > 0 in part a

[uncut tide]

ok

[torpid mirage]

So it's clear that's the behaviour at the right of zero

And the left is undefined

[uncut tide]

ok

makes sense now

[torpid mirage]

The hard part is over i think

[uncut tide]

nice

and when a=0 we literally have the graph 1/sqrtx

which is ok

[lone heart]

@uncut tide Has your question been resolved?

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[hybrid ravine]

how do I start this?

[charred jewel]

area of triangle = 1/2 |a| |b| sinθ

[charred jewel]

^

[hybrid ravine]

sorry idk what to do with that info😭

[charred jewel]

well ok can you first find what a dot a is

and b dot b

[hybrid ravine]

|a|^2 and |b|^2

[charred jewel]

and whats a dot b

[hybrid ravine]

|a||b|costheta

[charred jewel]

cool

[charred jewel]

now this is the formula for area of a triangle

and then you you have |a|² and |b|²

any ideas

[hybrid ravine]

root them?

[charred jewel]

no

[charred jewel]

you could try squaring this isnt it

[hybrid ravine]

oh

yes

[charred jewel]

if you square it what do you get

[hybrid ravine]

1/4|a|^2|b|^2sin^2theta

[charred jewel]

alright now you have a sin² in there

how would you relate that to the |a||b| cos theta

[hybrid ravine]

im gonna take a shot in the dark and say cos^2theta=1-sin^2theta?

[charred jewel]

other way around

sin² = 1 - cos²

[hybrid ravine]

right yep

[charred jewel]

so you get |a|²|b|²(1-cos²)

[hybrid ravine]

mhm

[charred jewel]

now can you progress with that?

[hybrid ravine]

i'll try ty

oh i got it ty again

.close

[lone heart]

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Use .reopen if this was a mistake.

[lone heart]

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Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[tough citrus]

Andi plays in a rectangular garden which has diagonal lengths of (3x + 2)m and (5x - 14)m. The length of the diagonal of the garden is...

[tough citrus]

My class has never been taught this

So I don't even know where to start

[candid widget]

diagonal lengths should be equal in a rectangle

so 3x+2 = 5x-14

solve for x, sub it in one of the equations

and boom

[tough citrus]

X is 8

The answer is literally not there

There's only
A. 7m
B. 17m
C. 25m
D. 26m

Oh wait

I have to calculate it lmao

.close

[lone heart]

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[lone heart]

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Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[jade comet]

.open

so ive made it this far but my brain aint working lads so im gonna need some help

[wraith stirrup]

what you supposed to find

[jade comet]

question is at the top

of the photo

A kite is formed by joining the points A(a, b) ,B(−1, 3) , C(x, y) and D(3, − 5) .
Determine the equations of the diagonals BD and AC of this kite.
Given a = − 5 and y = 4 , find the values of b and x .
Find the area of the kite (without the use of a calculator).

its year 9 maths

[wraith stirrup]

how old are you cuz im not from the us

but its not that difficult

[jade comet]

australia

im 15

[wraith stirrup]

so first

to find the diagonals

you should someting in form of: y = ax + b

you know that right ?

[jade comet]

yeah

we say y=mx+b

[wraith stirrup]

yeah its the same

so to find m you need 2 points

so for BD

the coordinates of B and D are given

m = (Y2-Y1)/(X2-X1)

[jade comet]

yes ive done the first bit

y=-2x+1
ive gotten to there

[wraith stirrup]

and where are you stuck

[jade comet]

the rest of it

[wraith stirrup]

you should try to form another equation example AB

[jade comet]

i been trying to do that

how would you approach the question

[wraith stirrup]

so i woul try to form other equations than the diagonal

so i can get some equations with a and b (coordinates of A) and try to find the values like this

[jade comet]

ill try fogire ot pit

figure it out

ok

right now for AC

I have C=(-3/2,4) and for A I only have (5,b)

Where do I go from there?

[honest prairie]

find the midpoint of BD

and the angle intersects at 90° at that point

so the angle would be -1/m

[jade comet]

im so fucking confused man

[honest prairie]

this is point BD

find the midpoint of BD and line AC intersects it at a right angle

@jade comet

[jade comet]

idk how to do that bro I aint been taught that shit

I know how to find midpoint

how do I find the line AC intersects with at a right angle

[honest prairie]

do you know how to find the gradient given 2 points?

[jade comet]

like uh

put it into a question

ill tell you

[jade comet]

yo how you draw that line btw I need that

[honest prairie]

m*(-1/m)

[honest prairie]

y=-2x+1

[honest prairie]

m in this case would be -2

so the right angled gradient (AC) would be 1/2

so you'll have y=1/2x+c

and because BD and AC intercept you can find c

[jade comet]

how is the right angled gradient 1/2

[honest prairie]

m * (-1/m) = -1

[jade comet]

yes

[honest prairie]

if the first m equals to -2

find the second m

[jade comet]

wtf is the second m

how do you find htat

[honest prairie]

m1 * (-1/m2) = -1

[jade comet]

I thought it was the same thing

[honest prairie]

m1 is the gradient for the first equation
m2 is gradient for second

[jade comet]

ok and how do you find the second gradient again

[honest prairie]

(-2) * (-1/m) = -1

wait

[jade comet]

bruh

[honest prairie]

lemme check something rq

[jade comet]

please

[honest prairie]

ok mb

its m * m = -1

-2 * 1/2 = -1

[jade comet]

bro this is what I got so far

im lost

*negative one

[honest prairie]

btw you forgot a bracket at the very top

=-2(x-(-1))

[honest prairie]

have you done part A with finding the equations of BD and AC?

[jade comet]

ty

[jade comet]

Idk 😭

im lost bro

[honest prairie]

you already know BD is y=-2x+1

lets find AC now

so you know that m = 1/2 right?

(for AC)

and it also intersects the midpoint

[jade comet]

hold on a sec

ok

[honest prairie]

because in a kite the angles intersect at the midpoint at right angles

[jade comet]

why does m=1/2 for ac again

I cant remember

[honest prairie]

M1 * M2 = -1

perpendicular lines

[jade comet]

ok

perpendicular lines

type shit

[honest prairie]

multipliying their gradients give -1

and -2 * 1/2 = -1

so m = 1/2

[jade comet]

ok

i understand

definniteky

[honest prairie]

youve got y=1/2x + c

do you know what point it'll intersect at?

[honest prairie]

(the midpoint of BD)

[jade comet]

(1,-1)

because perpendicular lines

type shit

[honest prairie]

yes

so youve got x=1 y=-1 and m=1/2

now its just this

which will end up giving y=x/2-3/2

[jade comet]

ok

can you put that into y=mx+b?

is that possible

[honest prairie]

this is in y=mx+b

[jade comet]

oh its confusing as shit to look at

[honest prairie]

y = 1/2x - 3/2

[jade comet]

ohhhh

y=1/2x-3/2

ok

I see

[honest prairie]

and then b) and c) should be fine

ping me if you need help with those two as well

[jade comet]

i probably will

but ill figure it out i hope

[honest prairie]

$$\text{BD}$$
$$y=-2x +1$$

$$\text{AC}$$
$$y = \frac{1}{2} x - \frac{3}{2}

[ocean seal]

Crissyboi_15
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[jade comet]

How do you find the value of b in the equation?

is b 1/2?

[honest prairie]

$$y=\frac{1}{2} x +b$$

[ocean seal]

Crissyboi_15

[honest prairie]

this?

[jade comet]

question b)

Find the value of b and x

Whats the value of b

[honest prairie]

ohh

so

$$y-y=m(x-x)$$

using this

'

[ocean seal]

Crissyboi_15

[honest prairie]

a=-5

and m = 1.2

1/2

[jade comet]

y-y1=m(x-x1)?

[honest prairie]

yeah

i didnt know how to put subscript on that

[jade comet]

point slope thing method

[honest prairie]

yea

[jade comet]

Could we use our working out to find the slope of AC to find B?

noooo nvm

thinking of wrong thing

[honest prairie]

do you get it?

[jade comet]

yes

[honest prairie]

$$y=\frac{1}{2} x-\frac{2}3$$
$$y-y_{1}=m(x-x_{1})$$
$$x_{1}=-5$$

[jade comet]

hold on

So we do the equation for AC first?

y=1/2x-2/3?

[ocean seal]

Crissyboi_15

[jade comet]

wait I got it

dont worry

[honest prairie]

yeah because AC uses a

$$y=\frac{1}{2} x-\frac{2}3$$
$$y-y_{1}=m(x-x_{1})$$
$$x_{1}=-5$$
$$\text{or}$$
$$y-b=\frac{1}2(x-(-5))$$

[jade comet]

ok so

[ocean seal]

Crissyboi_15

[jade comet]

the equation I have so far is

y-b=½ (x+5)

is that right

[honest prairie]

wait first we'll do it one by one

$$y=\frac{1}{2} x-\frac{2}3$$
$$a=-5$$
$$y=\frac{1}{2}(-5)-\frac{3}2$$

[jade comet]

why does y=4 on the first one

sorry

oh wait

uh

fuck

[honest prairie]

wait your confsuing me

wait

yeah

it is

[jade comet]

ok

so y=1/2x-2/3

[honest prairie]

then solve for x

ohhh wait

noo

[jade comet]

no

[honest prairie]

its wrong

its C(x,y)

here

[jade comet]

a=-5?

[honest prairie]

yes

[jade comet]

whats a

[ocean seal]

Crissyboi_15

[jade comet]

oh

that one

yes

[honest prairie]

$$\text{A(a,b) and C(x,y) use}$$
$$y=\frac{1}2 x - \frac{3}2$$

[jade comet]

ok

yes

[ocean seal]

Crissyboi_15

[honest prairie]

that took a while...

[jade comet]

y=4 rogjt

[honest prairie]

still learning

yea

[jade comet]

y=4

so from there

we can do the other equation right

y-b=½ (x+5)

4-b=1/2 (x+5)

[honest prairie]

nono this still uses this equation

[jade comet]

to find b?

[honest prairie]

yes

b is the y coordinate

a is the x coordinate

[jade comet]

i thought you used point slope

[honest prairie]

couldnt you just sub it into teh formula

[honest prairie]

i was wrong :/

i mean you could i think?

[jade comet]

yes

you could

??

[honest prairie]

but its just easier subbing it in directly

no need for 10 extra step[s

[jade comet]

wait lemme just work it out

on point slope

really quick

[honest prairie]

ight

i dont think you can use it with point slope

thats for finding the equation

we;'ve already got the eqution

so you can use point slope in this case

[jade comet]

4-b=½ (-3/2--5)
4-b=1/2(7/2)

is that right to start with

[honest prairie]

$$4-b=\frac{1}2 (-\frac{3}2-(-5)}$$

[ocean seal]

Crissyboi_15
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[jade comet]

yeah

[honest prairie]

what happened to the first x

no thats wrong

you dont do that

[jade comet]

I put 3/2 in from C

thats not what I do

wait

[honest prairie]

$$y-y_{1}=m(x-x_{1})$$

[ocean seal]

Crissyboi_15

[honest prairie]

you can only sub it into y1 and x1

[jade comet]

4-b=1/2(x-5)

[honest prairie]

you cant use point slope in this case

[jade comet]

what do I use instead

[honest prairie]

your trying to find x and y

$$\text{A(a,b) and C(x,y) use}$$
$$y=\frac{1}2 x - \frac{3}2$$

[ocean seal]

Crissyboi_15

[jade comet]

ok so

y=1/2x-3/2

[honest prairie]

$$\text{sub a=x= -5 and y=4}$$

[ocean seal]

Crissyboi_15

[honest prairie]

just remember that theyre for two different cases

[jade comet]

thats confusing me now

So I just do the equation Y=1/2x-3/2?

or what

[honest prairie]

$$\text{solve A(-5,y) and C(x,4) given }$$
$$y=\frac{1}2 x - \frac{3}2$$

[ocean seal]

Crissyboi_15

[honest prairie]

thats bascially it

[jade comet]

b is 4

isnt it

I mean it links up there

so im just assuming its 4

[honest prairie]

tahts the wrong equation

[jade comet]

fuck

theres too many equations in this fucking question

[honest prairie]

$$\text{BD}$$
$$y= -2x +1$$

$$\text{AC}$$
$$y=\frac{1}2 x - \frac{3}2$$

[ocean seal]

Crissyboi_15

[honest prairie]

$$\text{you would use AC when dealing with A(a,b) and C(x,y)}$$

[ocean seal]

Crissyboi_15

[jade comet]

Ok ok ok

so so

Y=1/2x-3/2

is the equation I have to solve?

[honest prairie]

yes

sub in x= -5

for A()

and then y=4 for C()

[jade comet]

fuck

okay

thats alll i needed to know I confused you a shit load

and myself too

[honest prairie]

😭

do you get it though?

like the concepts

[jade comet]

yes

somewhat

so its the same I did before

-5 = 1/2x - 3/2

[honest prairie]

A(-5,-4)
C(11,4)

[jade comet]

that queation ends up being -4?

ok

then I need to rearrange that shit again right

[honest prairie]

here this is bascialy your kite

[jade comet]

1/2b=-3/2+-5
would that be the equation

[honest prairie]

$$\frac{1}2 b = -\frac{3}2 + (-5)$$

[ocean seal]

Crissyboi_15

[honest prairie]

$$(4)=\frac{1}2 x - \frac{3}2$$

[jade comet]

yeah but you need to rearrange it to isolate x

or b

[ocean seal]

Crissyboi_15

[honest prairie]

$$b=\frac{1}2 (-5) - \frac{3}2$$

[ocean seal]

Crissyboi_15

[honest prairie]

here these are the two equations

[jade comet]

1/2b=-3/2+-5 isnt this how its meant to be??

[honest prairie]

i just subbed it in directly

$$\text{solve A(-5,y) and C(x,4) given }$$
$$y=\frac{1}2 x - \frac{3}2$$
$$\text{let y=4}$$
$$(4)=\frac{1}2 x - \frac{3}2$$
$$\text{let a=-5}$$
$$b=\frac{1}2 (-5) - \frac{3}2$$

[ocean seal]

Crissyboi_15

[honest prairie]

is this better?

[jade comet]

you just put it in the middle without rearranging the equation?

[honest prairie]

no you can still rearragnge

i just put it there to show steps

[jade comet]

how do you rearrange it

like how I did?

[honest prairie]

$$y=\frac{1}2 x - \frac{3}2$$
$$\text{let y=4}$$
$$(4)=\frac{1}2 x - \frac{3}2$$
$$\frac{1}2 x=4+\frac{3}2$$
$$x=8+3$$
$$x=11$$

[ocean seal]

Crissyboi_15

[honest prairie]

is this good?

[jade comet]

to solve for b

x=-5

[honest prairie]

yes

[jade comet]

its the same htingt?

[jade comet]

so b is -4???

[honest prairie]

yes correct

[jade comet]

so what does this solve

[honest prairie]

this is for C(x,y)

you were solving for A(a,b)

[jade comet]

wait I had it the other way aorund

oh fuck

[honest prairie]

$$A(-5,-4)$$
$$C(11,4)$$

[jade comet]

I thought C(3/2,4)

[ocean seal]

Crissyboi_15

[jade comet]

what is 3/2 again

from

like im fucking lost

this is what I have so far

[jade comet]

Wtf was the purpose of 3/2 then??

[honest prairie]

???

[jade comet]

Oh wait

I figured it out

fuck me I wrote it really weirdly

[honest prairie]

talk about using the midpoint as x1 and y1

like how you got the values

C(x,4)

[jade comet]

yeah

I was thinking weirdly when I wrote that

like

idk

I was trying to do something else

[honest prairie]

this isnt needed

[jade comet]

Too many formulas

[honest prairie]

you just need this

[honest prairie]

this is technically incorrect

wait

BD?

this already is BD

[jade comet]

BD

[honest prairie]

you need to find AC

[jade comet]

yes

where did I say BC

[honest prairie]

Now to find the equation for BD,

[jade comet]

yes

thats what I wrote?

wdym

[honest prairie]

which is incorrect

[jade comet]

OHH

[honest prairie]

BD is -2x+1

you are looking for AC

[jade comet]

here we go

I fixsed it

I fixed it

[honest prairie]

Now to find the equation for AC, we know that the gradient of the intersecting angles in a kite multiply to give -1. So -2*m=-1 meaning m=1/2

we have y=x/2+c for AC and now we just need to find c

we can find c because they intersect at the midpoint (1,-1)

so using y-y1=(1/2)(x-x1) we can find the equation for AC

[jade comet]

Where did we get =3/2 from again?

[honest prairie]

which ends up being y=1/2x-3/2

[honest prairie]

^ from here

$$y-y_{1}=m(x-x_{1})$$
$$m=\frac{1}2$$
$$y_{1}=-1$$
$$x_{1}=1$$

[ocean seal]

Crissyboi_15

[honest prairie]

$$\text{subbing in the values gives}$$
$$y=\frac{1}2 x - \frac{3}2$$

[ocean seal]

Crissyboi_15

[jade comet]

ima take a break and come back later tomorrow

its 1230 am right now

and I think my brain isnt working well

I should be able to finish this tomorrow pretty easisly

Thanks for the help my brain didnt want to work tonight and got easily confused

[honest prairie]

gl

im in australia too :P

if your done then close this channel so others can use it @jade comet

[jade comet]

Ayy

.close

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[sudden vapor]

Can someone help me get started on this problem

[sudden vapor]

I know I need to use trig identities but just not sure where to start

Pls ping so I see ur message thanks

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@sudden vapor Has your question been resolved?

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<@&286206848099549185>

https://tenor.com/view/hello-vro-gif-15207430103528002530

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[violet flare]

hey

[violet flare]

$(|x_{1}|+|x_{2}|...+|x_{d}|)^{2}<=d(|x_{1}|^{2}+|x_{2}|^{2}...+|x_{d}|^{2})$

[ocean seal]

Zouni

[violet flare]

any ideas on how to approach this?

initially coming from this question:

[sour verge]

It’s giving Cauchy-Schwarz no?

[violet flare]

you reckon?

[sour verge]

Idk a tingling sensation tells me that’s the way to go

[violet flare]

idea from here?

[sour verge]

Yeah well let v_i = 1 for every i and u_i = |x_i|

That might work

[violet flare]

that just explodes

what about the sum of one of the second series equals d??

[sour verge]

It’s finite it doesn’t blow up. What do you get adding 1 d times

[violet flare]

if v_{i} =1, don't we just get 1+1+1+1...

[sour verge]

Yeah, 1+1+1+1… d times

[verbal halo]

Hello

[violet flare]

ah yes that it what I was attempting to say earlier

[sour verge]

Okok

[violet flare]

hello

[verbal halo]

Any math genius’s in here that can teach basic algebra /geometry

[sour verge]

!occupied

[lone heart]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[violet flare]

in the middle of something

[verbal halo]

I need it to pass my math test

Okay

[violet flare]

try Kahn academy or organic chemistry tutor they have alot of good resources

[verbal halo]

Okay got you

[violet flare]

Is it suitable to just leave it as is after that? set parameters for Cauchy-schwarz and then that's it?

[sour verge]

Yeah you just have to say what u_i and v_i are and maybe just explain why you get the inequality you’re after from those and you’re good

[violet flare]

ok ty

would you mind looking at a previous result?

[sour verge]

Yes of course

Well Iwouldn’t mind I mean

[violet flare]

you can ignore the example vector stuff, it was just asked as another part of the question

[violet flare]

thank you

[sour verge]

It looks good to me

[violet flare]

ok thanks man, appreciate it

could I pick your brain for one more question @sour verge ?

[sour verge]

Yes no worries

[violet flare]

what do you reckon for this one?

More of the wording of this one, and generalisation required for the decomposition

[sour verge]

Do you know what the trace of a triangular matrix (one like is drawn in the problem) is?

[violet flare]

sum of the eigenvalues (along the diagonal)?

[sour verge]

Well just in general the trace of a triangular matrix is just the sum of the entries on the diagonal right?

[violet flare]

yes

[sour verge]

Ok, now use the fact that the trace of a product of matrices doesn’t change when you change the order of the multiplication.

Like trace(XYZ) = trace(ZXY) for instance

[violet flare]

I thought that was only for cyclical shifts?

tr(XYZ)=tr(ZXY)=tr(YZX) etc

[sour verge]

Yeah mb I mixed them up

But you can use it still

[violet flare]

ah ok ok

[sour verge]

Can you go from there?

[violet flare]

not sure where you're getting at so no

[sour verge]

Okok. Well we know what trace(U*AU) is right?

[violet flare]

right?

[sour verge]

So we know that trace(U* A U) = trace(U U* A) = …

[sour verge]

Yeah you’ll end up with that

[violet flare]

why is the result significant

[sour verge]

Because that’s how you show that trace(U* A U) = trace(A)

And moreover U* A U is triangular, so its trace is just the sum of the diagonal entries, which are the eigenvalues

[sour verge]

Along with U U* = I

[violet flare]

AH

I see

Is their a compact way to show U*AU is triangular

[sour verge]

It’s pretty much given from how they define the Schur decomposition

[violet flare]

it's just, what I have doesn't seem sufficient. Opinion?

[sour verge]

Well I wouldn’t say that the trace of U* A U is the sum of its eigenvalues as that is not a very useful. The trace of any matrix is the sum of its eigenvalues

I would also provide an explanation detailing why trace(A) = trace(U* A U)

[violet flare]

Ok thank you

so wha sentence would you reccomend?

[sour verge]

Here instead you could say that the trace of U* A U is just the sum of its diagonal entries, and its diagonal entries are the eigenvalues of A.

Also I just realized we don’t even need to mention that the matrix is triangular it just needs to be a square matrix

[violet flare]

ah ok good note

so you believe a more thorough description of why tr(A)=Tr(U*AU) will make it suitable?

[sour verge]

Yep, as I feel like this is the core of the problem

It’s not immediately evident that this is the case

[violet flare]

ok thank you very much

[sour verge]

No worries

[violet flare]

I thought it was weirdly presented

.close

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[alpine sable]

Question: the rank theorem states that the rank plus the the null space is equal to the number of columns, correct?

[alpine sable]

The question is related to the image but I don't need help with the image, just the question

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@alpine sable Has your question been resolved?

[alpine sable]

<@&286206848099549185>

[alpine sable]

<@&286206848099549185>

[alpine sable]

<@&286206848099549185>

[dark breach]

Close yes, but replace null space with nullity or dim(null space)

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[manic matrix]

Ik the answer is 0 but I’m unsure how to get there

[pseudo ice]

,rccw

[manic matrix]

I keep resulting in getting positive infinity because I’m dividing by a really small number?? Idk

[pseudo ice]

You're "dividing by a really small number" as you go towards -inf?

[manic matrix]

yeah??? idk?

i know im going wrong, i just simply dont understand how to do this problem

[pseudo ice]

Hmmmm, the difficulty of using "small" when negative numbers are involved

[manic matrix]

i hate going out to negative infinity cuz then i have no idea what to do

[pseudo ice]

Generally "small" is about size in absolute value, in this context

[manic matrix]

actually

you make a good point

below 0 "small" doesn't really make sense huh

[pseudo ice]

Yeaaaa it's not a great term for negative numbers, generally I try and distinguish when it comes to them, "small negative" and all

[manic matrix]

gotcha

well in that case i still got no idea what to do

[pseudo ice]

But yea, as you go towards -inf, you get large in absolute value

[manic matrix]

so instead, we're dividing by a really large number

therefore getting 0

am i right?

[pseudo ice]

That's basically it, you're dividing by a number that's large in absolute value, so in size 2/x gets smaller and smaller, so closer to zero

[manic matrix]

ill keep in mind negative infinity in denominator would be an absolute value

tyvm for that valuable information

i always seem to have very small parts of information that i miss when i learn which ends up fucking up my mojo

so this is extremely appreciated

🙏

[pseudo ice]

Awwww well to be a bit clear, if you have -inf in your denominator, then that's like having something very big in size in the denominator, so makes the fraction small in absolute value (in other words, closer to zero)

[pseudo ice]

And it's always like that

[manic matrix]

yeah i was just thinking it as if -inf was a really small number

so i was getting positive infinity

but now that youve pointed that out, it doesnt make sense so making it the absolute value makes sense

[pseudo ice]

Yep as above, that description of "small" gets really confusing when you think about negatives

[manic matrix]

yup x)

tyvm

.close

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[weak swift]

3d desmos only allows one {} per equasion

[weak swift]

i need to have 4 equasions be true

or combine them into one or 2 equasions

a very obvious one is {xtan126 < y < xtan18}

but beyond that im stuck

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@weak swift Has your question been resolved?

[lone heart]

@weak swift Has your question been resolved?

[weak swift]

hm

[lone heart]

@weak swift Has your question been resolved?

[weak swift]

i think i found a video on this

maybe

[weak swift]

no it wasnt on it but i figgued it out

.close

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[cobalt shadow]

so the solution i was given says this (in red) simplifies to c^2/4pi

im stuck on why the pi it's multiplied by and the bottom exponent are cancelled out

[wanton wigeon]

Can you explain simply

What you are not able to understand

[cobalt shadow]

why the exponent of pi in the denominator and the pi being multiplied by the fraction go away/are cancelled out

[hushed locust]

,, \frac{\pi^1}{\pi^2} = \pi^1 \cdot \pi^{-2} = \pi^{1-2} = \pi^{-1} = \frac 1{\pi}

[ocean seal]

cloud

[cobalt shadow]

wait so why doesnt multiplying the fraction by pi make the numerator c^2pi and the denominator 4pi^3

[hushed locust]

,, \pi = \frac{\pi}1

[ocean seal]

cloud

[cobalt shadow]

i see now, thank you!

.close

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[jade comet]

A trapezium is enclosed by the straight lines y = 0 , y = 3 , y = 7 − x , and x = k , where k is a constant.
Find the possible values of k given the trapezium has an area of 24 units2.

[candid widget]

try plotting it

and find the area of the trapezium in terms of k

and equate it to 24

[jade comet]

how would you do that

[candid widget]

ur x=k is wrong

[jade comet]

yeah

[candid widget]

its like this

u want this area

now try expressing that area in terms of k instead of using k=2

[jade comet]

idk how to do that

you got a formula or something brop

[candid widget]

well notice how that is made up of a rectangle

and a triangle

u can separate the shapes

and add their areas

[jade comet]

idk how to do that bro

ngl

i need some formula or some shit

[lone heart]

@jade comet Has your question been resolved?

[alpine sable]

so i've divided up the area you want

[jade comet]

I know the answers from the textbook I just have to find the working out

Which will be more reassuring now

[alpine sable]

the red area is a rectangle

[jade comet]

yes

[alpine sable]

do you know its base?

[jade comet]

2

[alpine sable]

height?

[jade comet]

3

so

area of it is 6

???

[alpine sable]

yes

now the triangle

base?

[jade comet]

3

[alpine sable]

height?

[jade comet]

3

[alpine sable]

so the area?

[jade comet]

4.5

[alpine sable]

great

[alpine sable]

so we have 6 and 4.5

add those to get the total area

[jade comet]

10.5

now where do we go from there

[alpine sable]

uh so that was k=2 apparently (oops)

the triangle area doesn't change thankfully

nor the height for the rectangle

one slight change is that the base of the rectangle is no longer 2

[jade comet]

what does it change to

4

[alpine sable]

close

we want the distance from k to 4

[jade comet]

isnt that just 2

what changes

[jade comet]

I dont understand what the ? is meant to represent

I just got to ask first of all though

what is the point of having k=2 in there

[alpine sable]

it was intended as an example i think

[alpine sable]

how would you find the distance between 4 and a number k

[jade comet]

k-4?

idk

[alpine sable]

ooh very close

that's correct when k is bigger than 4

otherwise it's 4-k

[alpine sable]

so coming back to this, the base of the rectangle is 4-k

[jade comet]

oh i didnt know there was a difference

[alpine sable]

it's because distance isn't allowed to be negative

[jade comet]

alright

so the base of the rectangle is 4-k

[alpine sable]

yep

height is still 3

and we know the area of the triangle is 4.5

do you know the area of the rectangle

[jade comet]

i dont know how to get the area of a rectangle when the base is 4-k

is it the same idea

just 4x3

[alpine sable]

it's still base x height

the base is 4-k

so you'd do 3x(4-k)

[jade comet]

so its 12-k

?

[alpine sable]

not quite

[jade comet]

do you have to expand the brackets

[alpine sable]

yes

[jade comet]

so

12-3k?

[alpine sable]

yep!

[jade comet]

idk bro

ok

[alpine sable]

that's the area of the rectangle

so adding together we get 16.5 - 3k is our trapezium area

which is 24

[jade comet]

ohhhhhhhh

so that will help us find out what k can be?

[alpine sable]

ideally yes

[jade comet]

so what would be the next step from there

[alpine sable]

we know that 16.5-3k = 24

can you solve this

[jade comet]

yeah

1 sec

k=-2.5

sound right?

[alpine sable]

yeah seems legit

is that the same as your answer?

[jade comet]

Yep!

fuck that makes so much more sense now that you simplified it

[alpine sable]

yay

[jade comet]

thats all I needed bro

Thank you sm

[alpine sable]

glad to be of service

[jade comet]

I know it takes me a bit for math

I specialise in english

I can write like a king but cant do math for shit and im still somehow on pathway to advanced

[alpine sable]

i suck at english lol

maybe if we averaged out i'd turn into a well-rounded person

[jade comet]

haha

thanks bro

@alpine sable that made it so much easier than how my teaches made it

thank you so mucj

[alpine sable]

dw abt it

[jade comet]

.close

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[alpine sable]

1. A triangle ABC is taken in the plane with ∠B = 40° and ∠C = 60°. For a point D taken on the side BC, let E be the perpendicular leg lowered from D to AB. |AC| = 2 (|AE|+|DC|)

If the condition is met, what is ∠DAC?

[lone heart]

@alpine sable Has your question been resolved?

[alpine sable]

Anyone can help?

[lone heart]

@alpine sable Has your question been resolved?

[meager karma]

This just looks like coordinate spam kind of problem tbh

[lone heart]

@alpine sable Has your question been resolved?

[plush needle]

show what you did

[lone heart]

@alpine sable Has your question been resolved?

[lone heart]

@alpine sable Has your question been resolved?

[lone heart]

@alpine sable Has your question been resolved?

[mortal magnet]

you here?

draw a diagram

then with this diagram, fill in the rest of the angles in the triangle

rewrite AC = 2(AE + DC) to 1/2 = AE/AC + DC/AC

because you filled in the rest of the angles, you should get that DAC = x, DAE = ||80 - x||, ADE = ||x + 10||, ADC = ||120 - x||
from here, tells you that DC/sin(DAC) = AC/sin(ADC)
rearrange this to get DC/AC = sin(DAC)/sin(ADC)
for AE/AC, that = AE/AD * AD/AC
AE/AD is a trig ratio and so = sin(ADE)
AD/AC = sin(ACD)/sin(ADC) due to the reason I just told you
so AE/AD * AD/AC = sin(ADE) sin(ACD) / sin(ADC)
together, you get the equation
1/2 = AE/AC + DC/AC
1/2 = sin(ADE) sin(ACD) / sin(ADC) + sin(DAC) / sin(ADC)
after subbing in the angles, you get a trig equation for which you solve for x

expand everything into sin x and cos x, then group like terms together, so that the equation becomes (numbers) sin x = (numbers) cos x
rewrite this to be tan x = (a big fraction)
simplify the fraction down to something of the form tan x = (sin(m)+sin(n))/(cos(m)+cos(n)) which through a trig identity is really tan((m+n)/2)

from this, you get the value of x

[alpine sable]

great approach but ı could not figure out how to solve the trigionometric equation

[lone heart]

@alpine sable Has your question been resolved?

[terse gull]

@alpine sable what’s the question

[terse gull]

Have you drawn it out?

[terse gull]

ah this

he’s already done it for you

what part are you struggling with

ping me when u need me

[alpine sable]

I tried but i could not figure out

If we plug x=10 in trigonometric equation we find tan 10

Maybr we can use tan 3x formula to prove what we have found in equation

But a geometric solution may be better if you could help

@terse gull

[terse gull]

Sorry looking now

[alpine sable]

If you can help me with trigonometric equation

I will appreciate it

[terse gull]

Can u show me on paper what you got so far please

[alpine sable]

Mentioned here

I got that far

[terse gull]

wait i’m busy rn

<@&286206848099549185>

if no one helps i’ll come back when i’m free

[alpine sable]

I need to find x value

Angle

[sweet wedge]

someone pls help

[left isle]

!occupied

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[tepid perch]

whats 1+1?

[crimson dawn]

Undefined

[waxen bison]

Average math persons humor

Cmon bro

Mot funny

Not

[crimson dawn]

Bro's the kind of person to say 2

[waxen bison]

Why is bro laughing

[rocky copper]

How is recovery’s problem not solved yet

[still herald]

o

[alpine sable]

ı solved it 2 days ago guys thanks for all your help

ı can share it if anyone wants

[alpine sable]

Hi

[alpine sable]

Funny

[odd sluice]

close the channel then ig

[lone heart]

@alpine sable Has your question been resolved?

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[fallow solar]

does green' thrm apply to this region R

[rigid smelt]

not yet with the region as a whole

[fallow solar]

wdym

[rigid smelt]

you can not apply green's theorem on R as a whole, but you can try to partition R into smaller regions where green's theorem is applicable

[lone heart]

@fallow solar Has your question been resolved?

[fallow solar]

like this ?

but if I sum them over the whole region will follow green's thrm right ?

[rigid smelt]

not quite

consider this case which is similar to your case

but less curvy

consider this partion where I segment the donut in half

can you see that we now have two region D_1 and D_2 where green's theorem is applicable?

[fallow solar]

but why is it not applicale on D

because the hole right ?

[rigid smelt]

yes

[fallow solar]

aha I see

alright

[fallow solar]

but in here things cancel out

so It's okay to say $\int_{C_{2}+C_{1}} \vec{F}.\vec{dr} = \int\int_{D}curl\vec{F}dA$

right ?

[ocean seal]

Hamdy Hisham

[rigid smelt]

in your case, it still cancels out

[fallow solar]

yup

I'm talking in general

[fallow solar]

this holds right?

[rigid smelt]

ehh I can't really say that in general, with only one hole? maybe

in general, I would just try to partition it myself to see

[fallow solar]

I think more complicated regions are out of my study anyways

just a check

it works here right ?

[rigid smelt]

what is the orientation of C_3?

[fallow solar]

clockwise

[rigid smelt]

yeah, still work

[fallow solar]

alright thnx

.close

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[marsh sedge]

How do I get the formula to get the terms for these sequences?

[rocky roost]

First one is just +5 , +4 alternating
So every odd term gets +5 and every even term gets +4

[marsh sedge]

i got that already, i just don't know how to put it in math language. sorry for not elaborating